Question: You have found the following ages (in years) of all 5 lions at your local zoo: $ 6,\enspace 2,\enspace 18,\enspace 6,\enspace 10$ What is the average age of the lions at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{6 + 2 + 18 + 6 + 10}{{5}} = {8.4\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $6$ years $-2.4$ years $5.76$ years $^2$ $2$ years $-6.4$ years $40.96$ years $^2$ $18$ years $9.6$ years $92.16$ years $^2$ $6$ years $-2.4$ years $5.76$ years $^2$ $10$ years $1.6$ years $2.56$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{5.76} + {40.96} + {92.16} + {5.76} + {2.56}} {{5}} $ $ {\sigma^2} = \dfrac{{147.2}}{{5}} = {29.44\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{29.44\text{ years}^2}} = {5.4\text{ years}} $ The average lion at the zoo is 8.4 years old. There is a standard deviation of 5.4 years.